Ice cream usually comes in 1.5 quart boxes (48 fluid ounces), and ice cream scoops hold about 2 ounces. However, there is some v
ariability in the amount of ice cream in a box as well as the amount of ice cream scooped out. We represent the amount of ice cream in the box as X and the amount scooped out as Y. Suppose these random variables have the following means, standard deviations, and variances. Mean SDVariance 48 0.25 0.0625 (a) An entire box of ice cream, plus 6 scoops from a second box is served at a party. How much ice cream do you expect to have been served at this party? fluid ounces 60 What is the standard deviation of the amount of ice cream served? (Round your answer to two decimal places.) 1.03 Xfluid ounces Enter a number (b) How much ice cream would you expect to be left in the box after scooping out one scoop of ice cream? That is, find the expected value of X - Y. fluid ounces What is the standard deviation of the amount left in the box? (Round your answer to two decimal places.) fluid ounces (c) Using the context of this exercise, explain why we add variances when we subtract one random variable from another Initially we do not know exactly how much ice cream is in the box, then we scoop out an unknown amount. Because we removed an unknown amount from a box where the amount of ice cream was unknown, we must add the variances to be confident of the difference when we subtract one unknown random variable from another unknown random variable Initially we do not know exactly how much ice cream is in the box, then we scoop out a known amount. Because we removed a known amount from a box where the amount of ice cream was unknown, we must add the variances to be confident of the difference when we subtract one known random variable from another unknown random variable. Initially we know exactly how much ice cream is in the box, then we scoop out an unknown amount. Because we removed an unknown amount, we must add the variances to be confident of the difference when we subtract one unknown random variable from another known random variable Initially we do know exactly how much ice cream is in the box, then we scoop out a known amount. Because we removed a known amount from a box where the amount of ice cream was known, we must add the variances to be confident of the difference when we subtract one known random variable from another known random variable
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Answer:
Explanation:
Mass of the cable car, m = 5800 kg
It goes 260 m up a hill, along a slope of
Therefore vertical elevation of the car =
Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = ). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.
Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.
Hence, total change in energy = mgh =
where, g = acceleration due to gravity
h = height/vertical elevation
Answer:
a)Initial speed of the projectile = 196.2 m/s
b)Maximum altitude = 490.5 m
c) Range of projectile = 3398.28 m
d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m
Explanation:
Time of flight of a projectile is given by the expression,
Here θ = 30° and t = 20 s
a)
Initial speed of the projectile = 196.2 m/s
b) Maximum altitude is given by
Maximum altitude = 490.5 m
c) Range of projectile is given by
Range of projectile = 3398.28 m
d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s
Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s
We have equation of motion s = ut + 0.5 at²
Horizontal motion
u = 169.91 m/s
a = 0 m/s²
t = 15 s
Substituting
s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m
Vertical motion
u = 98.1 m/s
a = -9.81 m/s²
t = 15 s
Substituting
s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m
Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m